By Sebastian Hayes

Unit Fractions

In an earlier post about Egyptian Fractions I discussed whether it was possible to present any (proper) fraction as a sum of unit fractions and posed the following questions :

(1) Can every proper fraction a/b (where a,b are positive integers) be expressed as a sum of unit fractions?

(2) For any given a/b, how many ways are there of representing it as a sum of unit fractions, supposing it to be possible ?

(3) Is there a way to reduce the number of terms in the series of unit fractions to a minimum, supposing a/b can be so represented ?

These are my investigations (undertaken without looking up the relevant article in Wikipedia which I recommend you do.)
Take two positive whole numbers a and b and assume that a/b is a ‘proper’ fraction, i.e. a < b as in 3/8 or 4/15The challenge is to find a method to transform a/b into a sum of unit fractions, i.e. fractions which all have 1 as numerator such as 1/6, 1/47 and so on.
In the trivial case where a = 1 we already have a ‘series’ of unit fractions, so we shall assume a ≠ 1. Also, we will assume that a/b is in its lowest terms, i.e. that a and b do not have a common factor greater than unity (note 1).
Nowa = 1 + (na – b)b n nb

(since if you multiply out you obtain (b + na – b) = na = a )nb nb b

For example, 7 = 1 + (5.7 ‒ 24) = 1 + 1124 5 5.24 5 120

Note that if na < b the numerator (na ‒ b)< 0 (is negative) and, since nb > 0 (i.e. is positive) this makes the whole second fraction negative and we end up with a subtraction, not an addition.
For example, if n = 2 we have (na ‒ b) = 2.7 ‒ 24 = ‒10 and

7 = 1 ‒ 10 = 1 ‒ 5 24 2 2.24 2 24

With the values we have chosen for a and b, it is clear that we need to have n at least 4, n ≥ 4Can we choose any value for n so long as it is greater or equal to 4 ? Seemingly, yes. For example, if n = 187

7 = 1 + (187.7 ‒ 24) = 1 + 1285 24 187 187.24 187 4488

However, we have still got the second fraction to reduce to a series of unit fractions and there may be more or less economical ways of doing this.
If we start with the lowest values of n that make an ‒ b positive, we find that, for a = 7, b = 24 we need to have n at least 4. In such a case

7 = 1 + (4.7 ‒ 24) = 1 + 4 = 1 + 124 4 4.24 4 4.24 4 24

So we have transformed 7/24 into a sum of just two unit fractions.
However, if we try out different values of a and b we will find that this was just a lucky fluke since, in general, the second fraction does not cancel down so readily. We can obtain the condition for this to happen, namely that (na – b) = n or b = (a ‒ 1)n
This gives the useful rule :

If b = (a ‒ 1)n then a = 1 + 1 …………..(i) b n b

We can check this by making up a/b in such a way that the condition is met. For instance, 65 = 5.13 and, according to (i),

6 = 1 + 1 which is the case.65 13 65

Similarly, since 51 = 3.17 we find that4 = 1 + 1 51 17 51

This situation is, however, highly unusual.
In general, given an arbitraryproper fraction a/b, and arbitrary n, we will have a unit fraction and a second fraction which is not unitary, i.e.a = 1 + c where c ≠1b n d

For example, with a = 7, b = 24, n = 5 we obtain

7 = 1 + (35 – 24) = 1 + 11 24 5 5.24 5 120

We can now express 11/120 as the sum of 1/n and c/d but this only pushes the problem further down the line. With n = 17

11 = 1 + (17.9 ‒ 120) = 1 + 33 120 17 17.120 17 2040

The second fraction is sufficiently small to be neglected in rough calculations but it is still there. And if we now express 33/240 as the sum of 1/n and e/f, we shall, unless we are very lucky, have a further non-unit fraction to deal with. Clearly, what we require is for the second fraction to disappear altogether, which will happen when (en — f) = 0, or f is an exact multiple of e, or when (en — f) = n in which case it will cancel with the n in the denominator (the case we have already considered). Can we, by a judicious choice of n, guarantee that this string of unit fractions will terminate either with zero or with a unit numerator?

Answer: It is always possible to express a proper fraction a/b as a terminating series of unit fractions. To do this, we only have to look for the smallest possible choice of n where an – b > 0 (positive). In the original case with a = 7, b = 24, it so happens that 4 is the smallest choice since 3 ×7 = 21 falls shortof 24 whereas 5 × 7 = 35 is too far above.
In the case of 7/24 we achieved our goal immediately which will not usually be the case, so let us look at another case to see how this checks out.7 = 1 + (3.7 – 16) = 1 + 516 3 3.16 3 48

Here, 3 is the smallest choice giving a positive second fraction. We continue :

5 = 1 + (10.5 – 48) = 1 + 248 10 10.48 10 480

nce again, we choose the smallest possible n. The second fraction reduces to 1/240 and if we hadn’t spotted that straight off we would obtain

2 = 1 + (2.240 ‒ 480) = 1 + 0480 40 240.480 240

We have thus expressed 7/16 as the sum of unit fractions

7 = 1 + 1 + 116 3 10 240

Is something like this going to occur if we systematically choose the smallest possible number for n compatible with (an ‒ b) > 0 ?In fact, yes. It all depends on the successive numerators of the non-unit fractions getting smaller each time. For example, 5 < 7 in

7 = 1 + 5 and 2 < 5 in 5 = 1 + 248 10 480 48 10 480

If the numerators always decrease, and are always positive integers, we will eventually arrive at a numerator which is either zero or unity. Can we show that this is necessarily the case ? Yes.
If a/b is a proper fraction, i.e. b > a, and b is not an exact multiple of a, then b/a will lie between (n ‒ 1) and n because we must have (an ‒ b) positive. Thus, b/a will be squeezed between successive integers or (n ‒ 1) < b/a < nThe distance between b/a and the smallest integer n such that n > (b/a) will thus always be less than unity, for example the distance between 16/7 and 3 is less than unity since 16/7 lies between 2 and 3. Now, since

(n ‒ b) < 1 means (na ‒ b) < aa

However, (na ‒ b) is the numerator of the second fraction in

a = 1 + (na – b) bb n nb

This shows that the numerator of the second fraction in the expansion is always less than the numerator we started with. Expanding the second fraction, we have the same situation

(na – b) = 1 + (m(na – b) ‒ nb) nb m mnb

Here, the numerator of the second fraction is less than the numerator we started with while the denominator is greater, since it has been multiplied by m. Thus, the last fraction in an expansion gets steadily smaller and, since the numerator is always a positive integer (or zero), we end up either with zero or a unit fraction.

More formally, is we define [b/a] as the smallest integer equal to or greater than b/a, we have the relation

a = 1 + ([b/a]a – b)b [b/a] nb

and, if we have ([b/a]a – b) > 0, then ([b/a] ‒ b/a) < 1 so

([b/a]a ‒ b) < a

Supposing ([b/a]a ‒ b) ≠ 1 we set ([b/a]a ‒ b) = c whence

c = 1 + [nb/c] ‒ nbnb [nb/c] [nb/c]nb

and we have ([nb/c] ‒ nb) < c < a and so on.

Also, note that we have assumed a < b which means that

na < nb so that (na – b) < (nb – b) or b(n – 1)

For example, in the case of 7 we took n = 4

a = 1 + (na – b)b n nb

(na – b) = 1 + m(na – b) – nb nb m mnb

Let (na – b) = c and let m be the least multiple of c < nbNow, mc < mb < a so the numerator is always diminishing and since 0 £ numerator we must eventually get back to 1 or 0.

Try another example : a = 8, b = 57, n = 8

8 = 1 + (64 – 57) = 1 + 7 c = 7 and 7 < 857 8 8.57 8 456

7 = 1 + 462 – 456 = 1 + 6 d = 6 < 7 < 8456 66 66.456 66 30096

6 = 1 + 30096 – 30096 = 1 . + 030096 5016 5016.30096 5016

Thus8 = 1 + 7 57 8 456

= 1 + 1 + 6 . 8 66 30096

= 1 + 1 + 1 . 8 66 5016

We conclude that it will always be possible to express a fraction as a sum of unit fractions. S.H.